Caluculate the molar solubility of PbCl2 in 0.0040 M KI solution. The Ksp of PbI2 - 7.2 x 10^-9. The answer should be 3.29 x 10^-4.
This is my work so far
PbI2<=====>Pb^+2 + I-
PbKI2----->Pb^+2 + KI^-1
7.2 x 10^-9=(s)(s+0.0040M)
Assume s<<0.0040
7.2 x 10^-9=(s)(0.0040M)
s=1.8 x 10^-6M
I don't get the right answer. My teacher told my class to assume to do the s<<the molar solubility given in the question. When you get an answer that's invalid then do a method called the successive approximation method. The way I tried to solve the question my s<< isn't invalid but my final answer isn't correct.
Did you really want to calculate the solubility of PbCl2 or did you mean to calculate the solubility of PbI2 in a KI solution? I suspect the latter.
I........solid.....0.......0
C........solid.....x.......x
E........solid.....x.......x
The latter is correct
You made a few errors at the beginning due to carelessness. This is a common ion problem (the common ion is I^-) and the addition of a common ion always decreases the solubility from what it would be in a pure water solution. Here's the scoop.
........PbI2 ==> Pb^2+ + 2I^(note yours)
I.......solid....0........0
C.......solid....s........2s
E.......solid....s........2s
Ksp PbI2 = (Pb^2+)(I^-)^2
Substitute your value for Ksp.
(Pb^2+) = s = solubility
(I^-) =2s from PbI2 and 0.004 from KI 2s + 0.004.
Ionization of KI is like this(100% ionized.
...........KI ==> K^+ + I^-
I........4E-3.....0.....0
C.......-4E-3....4E-3...4E-3
E.........0......4#-3...4E-3
So you plug in s for Pb^2+
Plug in 2s + 0.004 for I^-
If you assume s<<0.004 then you drop the 2s, solve for problem and I get 4.5E-4 M for solubility.
If, however, I do not throw out the 2s, I get a cubic equation which I solved and obtained an answer of 3.31E-4M for solubility which is quite close to the 3.29E-4 you quote as the correct answer. The small difference probably is the 3.29 number is by successive approximations and the 3.31 number is solving the cubic. I can't show you on this forum how to use successive approximations. You can go to Google and find explicit instructions. In my opinion, there is enough difference between 3.31E-4 and 4.5E-4 to warrant using 2s + 0.004 for I^-
Hope this helps.
This helped a ton thank you very much
Could u actually explain your calculations for solving this without dropping the s from (s+0.004) and the with dropping it. I'm not getting the same answers as u
7.2E-9 = (s)(2s+0.004)^2
Assume 2s + 0.004 = 0.004, then
7.2E-9 = (s)(0.004)^2
s = 7.2E-9/(0.004)(0.004) = 4.5E-4 mols/L.
For keeping 2s+0.004 as is I did this.
7.2E-9 = (s)(2s+0.004)^2. First, I expanded the (2s+0.004)^2 to get
4s^2 + 0.016s + 1.6E-5 and that times s gives 4s^3 + 0.016s^2 + 1.6E-5s = 7.2E-9. The final cubic then becomes
4s^3 + 0.016s^2 + + 1.6E-5s - 7.2E-9 = 0
I solved that cubic to get the 3.31E-4M.
If you want to do it by successive approximations here is my best effort. My eyesight isn't that good anymore and I don't always hit the right key. Even when I proof it I don't always catch an error but here goes. But I'm leaving the algebra for you; basically it's done as I wrote above for making the assumption that 2s + 0.004 = 0.004.
The idea behind successive approx is to get a first answer (which will be too large), plug that for 2s and solve again(which will be too small) and keep going back and forth until more approximations don't change the final answer enough to continue.
7.2E-9 = (s)(2s+0.004)^2
s-1 = 7.2E-9 - (0.004)^2 = 4.5E-4 as above.
That's the first answer; now plug that back for 2s like this.
7.2E-9 = (s)(9E-4 + 0.004)^2 =(s)(4.9E-3)^2 = (s)(2.4E-5) so solving for s = 7.2E-9/3.4E-5 = 3E-4 as the second try.Note that the first one is too high and the second one to low. The third one is 7.2E-9 = (s)(2*3E-4 + 0.004)^2 and I get 3.40E-4 (this one is large).
Fourth one is 3.29E-4
Fifth one is 3.32E-4
Sixth one is 3.31E-4 and I quit there since the answer only changed by 1 in 3rd place, technically 0.000331 and 0.000332 which is 1 in the 5th place. Successive approximations mehod is OK if you don't know how to solve a cubic but with computers it is easy to find a program on the web that will solve for cubic for you. That's what I did. You can see the effect of dropping the s from s+0.004 vs using it. 4.5E-4 vs 3.31E-4 is approximately 20% error. I don't usually do these problem with such detail but if you have trouble solving 7.2E-9 = (s)(0.004)^2 I'm trying to help you get started. I think you need to hone your algebra and calculator skills. My first guess, from years of doing this in the classroom, is that you forgot to square the 0.004. Cheers. Bob
Could u actually explain your calculations for solving this without dropping the s from (s+0.004) and the with dropping it. I'm not getting the same answers as u
You reposted you're same request; didn't you ready my 1:40 am response. I did what you wanted AND in GREAT detail. I worked what you wanted step by step. Did I do all that work for nothing? I made a typo eight lines up from bottom line. It should read "which is 1 in the 6th place" and not one in the 5th place.
To calculate the molar solubility of PbCl2 in a 0.0040 M KI solution, we first need to understand the reaction that occurs when PbCl2 dissolves in water. The balanced equation is:
PbCl2 (s) ↔ Pb2+ (aq) + 2Cl- (aq)
The Ksp expression for this reaction is:
Ksp = [Pb2+][Cl-]^2
Given that the Ksp of PbCl2 is 7.2 x 10^-9, we can set up an equilibrium expression:
7.2 x 10^-9 = [Pb2+]([Cl-]^2)
Now, let's consider the effect of adding KI to the solution. KI will provide an additional source of iodide ions (I-). However, it does not directly affect the concentration of lead ions (Pb2+).
Since we are given that the initial concentration of KI is 0.0040 M, we can assume that the concentration of iodide ions (I-) will be approximately equal to this value. Therefore, [I-] = 0.0040 M.
Now, we need to calculate the molar solubility of PbCl2, which is defined as the concentration of dissolved Pb2+ ions. Let's assume the molar solubility is "s" (in moles per liter).
Since PbCl2 dissociates into one Pb2+ ion and two Cl- ions:
[Pb2+] = s
[Cl-] = 2s
Substituting these values into the Ksp expression:
7.2 x 10^-9 = (s)(2s)^2
7.2 x 10^-9 = 4s^3
Now, solve for s:
s^3 = (7.2 x 10^-9) / 4
Taking the cubic root:
s = (7.2 x 10^-9)^(1/3) / 2^(1/3)
Calculating this expression, we find that s is approximately 3.29 x 10^-4 M, which matches the expected answer.
It's important to note that the assumption s << 0.0040 M was not necessary in this particular problem since the value of s obtained was much less than 0.0040 M. However, in cases where the obtained value of s is not significantly smaller than the initial concentration, the successive approximation method can be used to obtain a more accurate value.